Tentukanlah integral tentu berikut ini
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Integral Tertentu Dgn substitusi₀⁴∫ (2x +1) √(x + x²) dx =
u = x + x²
du = (2x +1) dx
u = x + x²
x= 4 --> u = 4 +16 = 20
x = 0 --> u = 0 + 0 = 0
₀⁴∫ (2x +1) √(x + x²) dx = ₀²⁰∫ √ u d u = ₀²⁰∫ u^(`1/2) du
= [ 2/3 u^(3/2)]²⁰₀
= [2/3 u√u ]²⁰₀
= (2/3) 20√20
= (2/3) 20 (2√5)
= 80/3 √5